Calculus 10th Edition

by
Larson, Ron; Edwards, Bruce H.

Chapter P - P.1 - Graphs and Models - Exercises: 62

Answer

The intersection points are $(x,y) = (-4,3)$ and $(x,y)=(-5,0)$.

Work Step by Step

The points of intersection are found by solving the system consisting of equations of the curves. In this problem we have
\begin{align}
&x^2+y^2=25\\
-&3x+y=15.
\end{align}
First express $y$ in terms of $x$ from the second equation:
$$y=15+3x.$$
Then we substitute what we expressed in the first equation
$$x^2+(15+3x)^2=25.$$ Expanding the bracket we get
$$x^2+225+90x+9x^2=25.$$
We can arrange all of the nonzero terms on the left and leave zero to the right to get the quadratic equation in $x$:
$$10x^2 + 90x +200=0.$$
We can divide the whole equation by $10$ to simplify:
$$x^2+9x+20=0.$$
Now the solutions are
$$x_{1,2} = \frac{-9\pm\sqrt{9^2-4\times20\times1}}{2} = \frac{-9\pm\sqrt{81-80}}{2} = \frac{-9\pm 1}{2}.$$
This evaluating for "plus" and for "minus" we get
$$x_1=\frac{-9+1}{2} = -4;$$
$$x_2=\frac{-9-1}{2}=-5.$$
Now to find the points we have to find the $y$ coordinates. Returning $x_1$ and $x_2$ in the expression that determines $y$ in terms of $x$ we have
$$y_1 = 3x_1+15= -12+15 = 3;$$
$$y_2=3x_2+15 = -15+15 = 0.$$
We get the points of intersection by arranging the $x$ and $y$ coordinates into pairs:
$$(x,y)\in\{(-4,3),(-5,0)\}$$